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Suppose $Z = X-Y$, $X,Y$ are independent, and $p_X = p_Y$. Then $\text{argmax} p_Z(z) = 0$ always?

For $p_Z$, I get $$ p_Z(z) = \int p_X(x)p_X(x-z) dx $$ and stuck at here.

Or would there be a counterexample?

In the case the above is not true in general, what if we assume $p_X$ is symmetric about a point?

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    Hint: Cauchy-Schwarz inequality. – 2017-02-27
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    How about this case，if X and Y are independent and N(0,1) distributed. – 2017-02-28
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    @Did Thanks for the hint! – 2017-02-28
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    @JGWang Yeah, how about it... Sorry but what is the point of your comment exactly? – 2017-02-28
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    @Did Thank you, you are right, I was confused by the "argmax". – 2017-03-01

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